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Prove that :-

[tex] \quad \leadsto \quad \bf \zeta ( z ) = \displaystyle \bf \sum_{\bf n=1}^{\infty} \bf \dfrac{1}{n^x} \cdot e^{-iy \cdot log ( n )} [/tex]

Where , z = x + yi

Riemann Hypothesis :) ​

Respuesta :

LammettHash

We have

[tex]e^{-iy \log(n)} = e^{\log(n^{-iy})} = n^{-iy}[/tex]

and

[tex]\dfrac1{n^x} \cdot e^{-iy \log(n)} = \dfrac{n^{-iy}}{n^x} = \dfrac1{n^{x+iy}} = \dfrac1{n^z}[/tex]

The sum

[tex]\displaystyle \sum_{n=1}^\infty \frac1{n^z}[/tex]

is exactly the definition of the Riemann zeta function [tex]\zeta(z)[/tex].

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